In mathematics<span>, a (real) </span>interval<span> is a set of real numbers with the property that any number that lies between two numbers in the set is also included in the set. For example, the set of all numbers x satisfying 0 ≤ x ≤ 1 is an </span>interval<span> which contains 0 and 1, as well as all numbers between them. So this should help you solve your problem.</span>
Answer:
I think it's 2,910 ft
Step-by-step explanation:
I did 6 times 97 times 5 to get the answer
Identity property of multiplication is used here because any factor multiplied by 1 is itself.
Answer:
Step-by-step explanation:
1. Find two numbers that add to make the coefficient of x (in this case, -5) and that multiply to make the constant term multiplied by the coefficient of x^2 (in this case, -2 x 3 = -6)
Two numbers that work are -6 and +1
-6 x +1 = -6
-6 + -1 = -5
2. Split the middle term into the two numbers that you found.
3x^2 -6x +x -2 = 0
I've put the -6 on the left side because in our next step, when we factorise, it will be easier than having the numbers the other way around.
3. Factorise the left side by taking out common factors from each pair. The pairs I'm talking about here are '3x^2 and -6x', and 'x and -2'
3x (x-2) +1 (x-2) = 0
4. You now have two numbers both being multiplied by the term x-2. We can rearrange this equation to give us two brackets being multiplied by each other.
(3x + 1) (x-2) = 0
5. According to the Null Factor Law, if two terms are multiplied together and the result is 0, then one of those terms must be 0. Make both terms equal to 0 and solve each for x.
3x + 1 = 0 x-2 = 0
3x = -1 x = 2
x = -1/3
6. The solutions to this equation are x = 2 and x = -1/3
Let the number be x
then product of number and 17 would be written as 17x
